Description: This tutorial teaches you how to upload images into a mysql database using PHP. Even though it sounds complicated, it is fairly simple and has many practical applications. One example of an application would be Forum User Images. Take user avatars, for example. It's impractical to upload avatar files to one common folder, because chances are sooner or later two users will have the same name for an avatar, and either one avatar will be overwritten, or the other not accepted, causing trouble. Image Databasing solves this problem by inserting the image data into its own unique row in a table, each assigned with an ID number instead of a filename.

Images can then be called from the database and be view using one php file for all images. How are they inserted into the database? By converting the data to base64. If you're confused, please bear with me, you will understand it soon.
There will be 3 PHP files in this tutorial:

readdir.php - this puts all the images in a folder into the database.

image.php - the actual image script that displays the image.

view.php - an example file that shows you how to call the image.

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Language: PHP Highlight Mode: PHP Last Modified: February 27th, 2009

Creating the Image Database
First, create a mysql database called base64imgdb
(this is the name that will be used throughout the tutorial)
Second, create a table called images with two rows. Name the
first one imgid, and give it the parameters
TYPE: INT EXTRA: auto_increment, and check the circle under
Primary. Name the second sixfourdata, and make it
TYPE: LONGTEXT. Here is the sql code:
CREATE TABLE `images` (
`imgid` INT NOT NULL AUTO_INCREMENT ,
`sixfourdata` LONGTEXT NOT NULL ,
PRIMARY KEY ( `imgid` )
);
The READDIR.PHP script
This script reads a directory within the server, selects all
the jpg and gif images, encodes them into base64, and uploads
them to the database, except in a different order. This is
because the script reads each image in a loop, and we would
like to keep a constant connection to the mysql database
instead of creating multiple ones. Here is the database
connection where username and password need to be changed:
&LT;?PHP$dbcnx=@MYSQL_CONNECT("localhost","username","password");IF(!$dbcnx){ECHO("<p>connection to database server failed!</p>");EXIT();}IF(!@MYSQL_SELECT_DB("base64imgdb")){ECHO("<p>Image Database Not Available!</p >");EXIT();}?>
Next we need to open the directory, where "./"is the directory
the readdir.php file is located:
$path = "./";
$dir_handle = @opendir($path) or die("Unable to open directory $path");
This is the hardest part of the script: sorting the image types,
reading the data using fopen, converting it using base64_encode,
and then inserting it into the table.
&LT;?PHPWHILE($file=READDIR($dir_handle)){$filetyp=SUBSTR($file,-3);IF($filetyp=='gif' OR $filetyp=='jpg'){$handle=FOPEN($path."/".$file,'r');$file_content=FREAD($handle,FILESIZE($path."/".$file));FCLOSE($handle);$encoded=CHUNK_SPLIT(BASE64_ENCODE($file_content));$sql="INSERT INTO images SET sixfourdata='$encoded'";@MYSQL_QUERY($sql);}}?>
This is the last and final part of the readdir.php: closing the
directory and stating the process is complete:
&LT;?PHPCLOSEDIR($dir_handle);ECHO("complete");?>
The Image Reader IMAGE.PHP
This file may be the hardest file to understand whenever you see
how simple view.php is, but bear with me, your patience will pay
off. This file takes a request, requests the row in the table,
decodes the data, and presents itself as an image. First, we
have to connect to the database again:
&LT;?PHP$dbcnx=@MYSQL_CONNECT("localhost","username","password");IF(!$dbcnx){ECHO("
connection to database server failed!
");EXIT();}IF(!@MYSQL_SELECT_DB("base64imgdb")){ECHO("
Image Database Not Available!
");EXIT();}?>
Now we need to find out which row it's requesting, which is done
using image.php?img=x:
&LT;?PHP$img=$_REQUEST["img"];?>
After this, we need to connect to the table, get the data, and
set it into variables:
&LT;?PHP$result=@MYSQL_QUERY("SELECT * FROM images WHERE imgid=".$img."");IF(!$result){ECHO("
Error performing query: ".MYSQL_ERROR()."
");EXIT();}WHILE($row=MYSQL_FETCH_ARRAY($result)){$imgid=$row["imgid"];$encodeddata=$row["sixfourdata"];}?>
Now here is the last and most confusing part of the file:
&LT;?PHPECHOBASE64_DECODE($encodeddata);?>
Now let me explain this. All this does is decodes the base64-encoded
image data, end echos it. That's it, nothing else.
VIEW.PHP (example viewer)
Okay, so you made it this far already. This is now the easiest
to copy and paste but hardest part to understand, where
image.php?img=1 matches with whatever row the image is on, for
example if it's row 357 then you would need to put
image.php?img=357:
<img src='image.php?img=1' border="0" alt="">
Now that wasn't so hard was it? But most of you are probably
wondering why when you link to a page, you get an image. This
is the reason: images arent defined by their 3 letter suffixes
(such as jpg or gif), but by how their headers are written.
IMAGE.PHP simply echos the image data, and acts like an image
even though it just proccesses the request. This is why you get
an image.
readdir.php:
&LT;?PHP################################ DB CONNECTION# CHANGE THESE VALUES ###############################$dbcnx=@MYSQL_CONNECT("localhost","username","password");IF(!$dbcnx){ECHO("
connection to database server failed!
");EXIT();}IF(!@MYSQL_SELECT_DB("base64imgdb")){ECHO("
Image Database Not Available!
");EXIT();}$path="./";$dir_handle=@OPENDIR($path) or DIE("Unable to open directory $path");WHILE($file=READDIR($dir_handle)){$filetyp=SUBSTR($file,-3);IF($filetyp=='gif' OR $filetyp=='jpg'){$handle=FOPEN($file,'r');$file_content=FREAD($handle,FILESIZE($file));FCLOSE($handle);$encoded=CHUNK_SPLIT(BASE64_ENCODE($file_content));$sql="INSERT INTO images SET sixfourdata='$encoded'";@MYSQL_QUERY($sql);}}CLOSEDIR($dir_handle);ECHO("complete");?>
image.php:
&LT;?PHP$dbcnx=@MYSQL_CONNECT("localhost","username","password");IF(!$dbcnx){ECHO("
connection to database server failed!
");EXIT();}IF(!@MYSQL_SELECT_DB("base64imgdb")){ECHO("
Image Database Not Available!
");EXIT();}$img=$_REQUEST["img"];$result=@MYSQL_QUERY("SELECT * FROM images WHERE imgid=".$img."");IF(!$result){ECHO("
Error performing query: ".MYSQL_ERROR()."
");EXIT();}WHILE($row=MYSQL_FETCH_ARRAY($result)){$imgid=$row["imgid"];$encodeddata=$row["sixfourdata"];}ECHOBASE64_DECODE($encodeddata);?>
view.php:
<html>
<body>
..
<img src='image.php?img=1' border="0" alt="">
..
</body>
</html>

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